-0.3x^2+0.1=-0.01

Solution for -0.3x^2+0.1=-0.01 equation:

-0.3x^2+0.1=-0.01

We move all terms to the left:

-0.3x^2+0.1-(-0.01)=0

We add all the numbers together, and all the variables

-0.3x^2+0.11=0

a = -0.3; b = 0; c = +0.11;
Δ = b2-4ac
Δ = 02-4·(-0.3)·0.11
Δ = 0.132
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{0.132}}{2*-0.3}=\frac{0-\sqrt{0.132}}{-0.6}
=-\frac{\sqrt{}}{-0.6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{0.132}}{2*-0.3}=\frac{0+\sqrt{0.132}}{-0.6}
=\frac{\sqrt{}}{-0.6}
$